Problem: Simplify and expand the following expression: $ \dfrac{p - 7}{3p + 6}-\dfrac{4p}{5p - 3} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3p + 6)(5p - 3)$ Multiply the first term by $\dfrac{5p - 3}{5p - 3}$ $ \begin{align*} \dfrac{p - 7}{3p + 6} \times \dfrac{5p - 3}{5p - 3} & = \dfrac{(p - 7)(5p - 3)}{(3p + 6)(5p - 3)} \\ & = \dfrac{5p^2 - 38p + 21}{(3p + 6)(5p - 3)}\end{align*} $ Multiply the second term by $\dfrac{3p + 6}{3p + 6}$ $ \begin{align*} \dfrac{4p}{5p - 3} \times \dfrac{3p + 6}{3p + 6} & = \dfrac{(4p)(3p + 6)}{(5p - 3)(3p + 6)} \\ & = \dfrac{12p^2 + 24p}{(5p - 3)(3p + 6)}\end{align*} $ Now we have: $ = \dfrac{5p^2 - 38p + 21}{(3p + 6)(5p - 3)} - \dfrac{12p^2 + 24p}{(5p - 3)(3p + 6)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{5p^2 - 38p + 21 - (12p^2 + 24p)}{(3p + 6)(5p - 3)} $ $ = \dfrac{5p^2 - 38p + 21 - 12p^2 - 24p}{(3p + 6)(5p - 3)} $ $ = \dfrac{-7p^2 - 62p + 21}{(3p + 6)(5p - 3)}$ Expand the denominator: $ = \dfrac{-7p^2 - 62p + 21}{15p^2 + 21p - 18}$